Let $a',b',c'$ be the length of angle bisectors from angle $A,B,C$ in $\Delta ABC$ respectively. Find area of $\Delta ABC$ in terms of $a',b',c'$.
My approach: $\Delta=1/2*a*a'*\sin(A/2+C)=1/2*b*b'*(B/2+C)=1/2*c*c'* \sin(C/2+A) a/ \sin A=b/ \sin B=c/ \sin C = abc/2\Delta$ then I don't know how to simplify. Or can it be simplified?
